Solve the problem.
-In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh
More than 180 pounds? Assume that we want 96% confidence that the error is no more than 4 percentage points.
A) 317
B) 501
C) 232
D) 658

Question

Quizplus · Accepted Answer

We need to use the formula for the sample size of a proportion with normal approximation:

$$n=\frac{z^2\cdot p\cdot (1-p)}{E^2}$$

where: 
- $z$ is the z-score corresponding to the desired confidence level. In this case, the confidence level is 96%, so $z=2.05$ (we use a standard normal distribution table or calculator to find this value).
- $p$ is our estimate of the proportion of people who weigh more than 180 pounds. We don't have a good guess for this, so we use 0.5 (which maximizes the sample size for a given error).
- $E$ is the maximum allowed error, which is 4 percentage points or 0.04.

Plugging in these values, we get:

$$n=\frac{(2.05)^2\cdot 0.5\cdot (1-0.5)}{(0.04)^2} \approx 658$$

Therefore, we need to survey at least 658 people to estimate the percentage of people who weigh more than 180 pounds with 96% confidence and a maximum error of 4 percentage points. The closest answer choice is (D) 658.

Solve the Problem. -In a Certain Population, Body Weights Are Normally Distributed with Normally